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March 26, 2017

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I need to create a 100ml buffer with a pH of 4.00 using 0.100M Benzoic acid (pKa = 4.20)and 0.180M sodium benzoate. What volume of each material do I need to make this buffer? I think I need to use pH=pKa + log(A/HA)but am not sure if that is correct. Any help would be appreciated.

  • biochem - ,

    Yes, the Henderson-Hasselbalch equation is what you use.
    If x = mL sodium benzoate (base)
    then 100-x = mL benzoic acid(acid).

    4.00 = 4.20 + log(0.18x)/[(100-x)*0.1]
    and solve for x

  • biochem - ,

    I think that is where I keep getting stuck, with the basic math. I have 0.631 = .180x/10-0.1x and don't know where to go from there. Math was a long time ago for me :)

  • biochem - ,

    0.631 = (0.180x)/[(100-x)*0.100]
    0.631*[(100-x)*.100] = 0.180x
    multiply 0.631*0.100 first to obtain 0.0631, then
    0.0631*(100-x) = 0.180x
    6.31-0.0631x = 0.180x
    6.31 = 0.0631+0.180x
    6.31 = 0.2431x
    x = 6.31/0.2431 = 25.96 mL
    100-x = 74.04 mL
    These should be rounded to 26.0 mL and 74.0 mL.
    You should check this by multiplying 26.0x0.180 = ?? and 74.0 x 0.100 mL = ??, substitute into the HH equation and see if you arrive at a pH of 4.00.

  • biochem - ,

    thank you so much. I just recently went back to school and havent had chemistry in more than 4 years and math has been even longer. I really appreciate the help :)

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